Finding the shortest tunnel from a point to a line in three dimensions

Key takeaway: This is a vector geometry problem involving a line in 3D, projection ideas, and distance comparison in a realistic modelling context.

Part (a): vector equation of the line PQ

The points are

$P(-300,400,-150), \qquad Q(300,300,-50).$

A vector along the line is found by subtracting coordinates:

$\overrightarrow{PQ}=Q-P=(600,-100,100).$

So a vector equation for the line is

$\mathbf r = \begin{pmatrix}-300\\400\\-150\end{pmatrix} + \lambda \begin{pmatrix}600\\-100\\100\end{pmatrix}.$

Any nonzero scalar multiple of the direction vector is also acceptable.

Part (b): shortest tunnel from M to the pipeline

The plane is

$2x+3y-5z=300,$

and the point on it is $M=(100,k,100)$. Since $M$ lies on the plane,

$2(100)+3k-5(100)=300.$

That gives

$200+3k-500=300 \Rightarrow 3k=600 \Rightarrow k=200.$

So $M=(100,200,100)$.

The shortest tunnel from a point to a line is the perpendicular from the point to the line. Let a general point on the pipeline be

$R=(-300,400,-150)+t(600,-100,100).$

Then

$R=(-300+600t,\ 400-100t,\ -150+100t).$

For the shortest distance, $\overrightarrow{MR}$ must be perpendicular to the direction vector $(600,-100,100)$. So

$(M-R)\cdot (600,-100,100)=0.$

Compute

$M-R=(400-600t,\ -200+100t,\ 250-100t).$

Dotting with the direction vector gives

$(400-600t)600+(-200+100t)(-100)+(250-100t)100=0.$

This simplifies to

$255000-370000t=0 \Rightarrow t=\frac{51}{74}.$

Substitute back to find the meeting point:

$R=\left(-300+600\cdot\frac{51}{74},\ 400-100\cdot\frac{51}{74},\ -150+100\cdot\frac{51}{74}\right).$

So the tunnel meets the pipeline at

$\left(\frac{5700}{37},\ \frac{24550}{74},\ -\frac{1950}{37}\right)$

which can also be written in simplified decimal form if desired.

The tunnel length is the distance $MR$. Using the vector difference

$M-R=\left(100-\frac{5700}{37},\ 200-\frac{24550}{74},\ 100+\frac{1950}{37}\right),$

we obtain the exact length

$|MR|=\frac{50\sqrt{370}}{37}\text{ m}.$

Part (c): compare with the existing accessways

Compute the lengths of the existing accessways:

$OP=\sqrt{(-300)^2+400^2+(-150)^2}=\sqrt{272500}=50\sqrt{109},$

$OQ=\sqrt{300^2+300^2+(-50)^2}=\sqrt{182500}=50\sqrt{73}.$

The new tunnel has length

$\frac{50\sqrt{370}}{37}\approx 26.0\text{ m},$

while

$50\sqrt{73}\approx 427.2\text{ m},\qquad 50\sqrt{109}\approx 522.0\text{ m}.$

So the new accessway is far shorter than both existing ones, and the company should build it if the model is accepted.

Part (d): one limitation of the model

One limitation is that the model treats the pipeline and mountain side as perfect geometric objects. Real terrain is uneven, and a tunnel may not be a straight segment through solid rock. The model also ignores construction constraints such as safety regulations, soil stability, and whether the tunnel can actually be drilled along the mathematically shortest path.

A strong way to handle this kind of question is to separate the geometry from the context: first use vectors to find the perpendicular distance, then interpret whether the result is practically worthwhile.

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Pitfalls the pros know 👇 A common mistake is to think the shortest tunnel should meet the pipeline at the midpoint of PQ or at the point on the line closest to the origin. Neither is generally true. The shortest segment from a point to a line is perpendicular to the line, so the direction vector must be orthogonal to the tunnel vector. Another frequent error is forgetting to use the fact that M lies on the plane before solving for k. That condition is essential, and if you skip it, the rest of the geometry starts from the wrong point. Finally, when comparing lengths, keep units consistent and compare actual distances, not coordinate values.

What if the problem changes? If the point were changed to $M(100,k,80)$ while still lying on the same plane, the value of $k$ would change and the foot of the perpendicular on the line would move accordingly. The method would be identical: first enforce the plane equation to determine the missing coordinate, then use a dot-product perpendicularity condition to find the closest point on the line. If the pipeline were instead described by a different line, for example one parallel to $PQ$ but shifted in space, the shortest tunnel would still be found by projection ideas, but the actual meeting point and length would change.

Tags: vector equation of a line, perpendicular distance in 3D, dot product orthogonality